An A level text book claims that one can find the quotient by first:

1.) Setting up an identity, $f(x)≡ Q(x)(divisor) + remainder$

2.) Finding the coefficients

However, another A level text book says, "Note. This theorem gives a (simple) method for evaluating the remainder only. If the quotient is required, long division must be used."

The question is: Divide $x^3 + x^2 - 7$ by $x-3$ using the remainder theorem.

In this example,

1.) They set up the identity: $x^3 + x^2 - 7 ≡ (Ax^2 + Bx + C)(x-3) + D$

2.) They let $x=3$ to find coefficient $D$

3.) They let $x=0$ to find coefficient $C$

4.) To find coefficients $A$ and $B$, the text book then goes on to "comparing the coefficients". No more detail is given as to how "comparing the coefficients" to find $A$ and $B$ is achieved.

Can you find coefficients $A$ and $B$ using this method ONLY? If so, how?

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    $\begingroup$ Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. $\endgroup$ – Zev Chonoles Jul 21 '13 at 17:28
  • $\begingroup$ @Zev Chonoles Thanks $\endgroup$ – PurpleJess Jul 21 '13 at 17:37
  • $\begingroup$ @ThomasAndrews Sorry, yes, I meant that. I have edited it now. $\endgroup$ – PurpleJess Jul 21 '13 at 17:47

We write $$x^3 + x^2 - 7 ≡ (Ax^2 + Bx + C)(x-3) + D.\tag{1}$$ Put $x=3$. That conveniently kills all but $D$ on the right. Substituting on the left, we get that $D=3^3+3^2-7=29$. We can continue making other simple substitutions, and get linear equations that determine our coefficients.

But you asked about comparing coefficients. That is done as follows. Expand the right-hand side of (1). So in particular, multiply out $(Ax^2 + Bx + C)(x-3)$. We get after a while that the right-hand side is equal to $$Ax^3 +(-3A+B)x^2 +(-3B+C)x -3C+D.$$

This has to be the same polynomial as $x^3+x^2-7$.

So the coefficients of $X^3$ must be the same. That means $A=1$.

The coefficients of $x^2$ must be the same. That means $-3A+B=1$.

The coefficents of $x$ must be the same. That means $-3B+C=0$.

And the constant terms must be the same. That means $-3C+D=-7$.

Now solve. We already know $D=27$ a "quickie" way. The rest of the coefficients are now easy to pick up.

  • $\begingroup$ Thanks! I didn't understand the comparing of coefficients part. I didn't realise that you had to multiply them out. Thank you so much for your response! $\endgroup$ – PurpleJess Jul 21 '13 at 18:03
  • $\begingroup$ You are welcome. You will bump into this again in integration, and elsewhere. There are in your situation other ways to get the coefficients, such as polynomial division, or appropriate substitutions, $\endgroup$ – André Nicolas Jul 21 '13 at 18:08

Assuming the equation was meant to be:

$$x^3+x^2-7 = (Ax^2+Bx+C)(x-3)+D$$

Setting $x=3$, we sse that $3^3+3^2-7 = 29 = D$.

Setting $x=0$, we see that $-7 = D-3C=29-3C$, so $C=12$.

Setting $x=1$, we see that $-5=(A+B+12)(-2)+29$, or $A+B+12 = 17$.

Setting $x=2$, we see that $5 = (4A+2B+12)(-1)+29 = 4A+2B+12 = 24$.

So now you have two linear equations for $A,B$, and you can solve those.

(Definitely redo my arithemetic, it could be in error.)

  • $\begingroup$ $A$ is obvious by comparing coefficients of $x^3$ - which can only arise in one way on the right-hand side. $\endgroup$ – Mark Bennet Jul 21 '13 at 17:58
  • $\begingroup$ Thank you! Embarrassingly obvious now that I've read your answer. $\endgroup$ – PurpleJess Jul 21 '13 at 18:00
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    $\begingroup$ @MarkBennet Sure, just trying to complete the problem as requested, not by the way I'd do it. :) $\endgroup$ – Thomas Andrews Jul 21 '13 at 18:00
  • $\begingroup$ Indeed - but learning how to do this stuff, I always wanted to know the quickest way of solving. However, reliable systematic methods are also important. $\endgroup$ – Mark Bennet Jul 21 '13 at 18:04

(I am not completely sure that this is what you are looking for.)

Let $$p(x) = x^3 + x^2 - 7 = (Ax^2 + Bx + C)(x-3) + D.$$

You want to find $A,B,C,$ and $D$. Note first that you are dividing by $x-3$, so the remainder of the division will be a constant. That is we have $$\begin{align} {\color{blue} 1}x^3 + {\color{red} 1}x^2 + {\color{gray} 0}x + {\color{green}{-7}} &= {\color{blue}A}x^3 + {\color{red}B}x^2 + {\color{gray}C}x + {\color{red}{-3A}}x^2 + ({\color{gray}{-3B}})x + ({\color{green}{- 3C + D}})\\ &= {\color{blue}A}x^3 + ({\color{red}{B - 3A}})x^2 + {\color{gray}{(C - 3B)}}x + ({\color{green}{- 3C + D}}) \end{align} $$ Now, you know that two polynomials are equal if and only if the coefficients match up. That is, we must have $$\begin{align} 1 &= A\\ 1 &= B - 3A\\ 0 &= C - 3B \\ -7 &= -3C + D. \end{align} $$ You already have $A$, so from the second equation you can find $B$. That gives $C$ from the third equation. And so from the last equation you have $D$.

  • $\begingroup$ @PurpleJess: Glad to help. $\endgroup$ – Thomas Jul 21 '13 at 18:02

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